# Radial length contraction = mass anisotropy?

1. Mar 8, 2013

### Vincentius

Mach's principle is considered falls by many because mass anisotropy would not be consistent with the Huges-Drever experiments, which showed the isotropy of nuclear resonance. However, anisotropy appears as well in GR in the form of radial length contraction, which can be interpreted as increase of inertia in the radial direction (the observer at infinity can not tell the difference). Thus one would expect anisotropic behavior of mass in GR just the same as according to Mach's principle. The fact that GR time dilation is isotropic does not change this anisotropic aspect of GR. So why is it an issue within Mach's principle and not within GR?

I tried to find a resource on a GR harmonic oscillator, but I can not find a paper that answers this question really.

2. Mar 8, 2013

### Staff: Mentor

Can you give a reference for this? I'm not sure what you're referring to.

3. Mar 8, 2013

### Staff: Mentor

I echo Peter Donis' comment. This needs a mainstream scientific reference.

4. Mar 8, 2013

### bcrowell

Staff Emeritus
Hughes-Drever experiments don't test Mach's principle, they test Lorentz invariance. See Will, "The Confrontation between General Relativity and Experiment," section 2.1.2, http://relativity.livingreviews.org/Articles/lrr-2006-3/fulltext.html [Broken] . There is also a WP article: http://en.wikipedia.org/wiki/Hughes–Drever_experiment As both of these sources explain, the original motivation had to do with Mach's principle, but that turned out to be wrong.

Mach's principle can hold while Lorentz invariance also holds, as in Brans-Dicke gravity. Mach's principle can fail while Lorentz invariance holds (GR). I think most Lorentz-violating test theories do not embody Mach's principle, since they're typically designed to be minimal variations on SR/GR. So I don't think there's any close logical relationship between the two. Cocconi and Salpeter were working in 1958, long before BD gravity had clarified these issues by providing a viable Machian test theory.

No, Mach's principle (as embodied in BD gravity) is really not viable due to solar system tests. If Mach's principle had been falsified by Hughes-Drever in 1960, then BD gravity would never have been a viable, Machian theory.

No, length contraction does not create anistropy. GR is locally equivalent to SR, which is an isotropic theory.

Last edited by a moderator: May 6, 2017
5. Mar 8, 2013

### Naty1

bcrowell posts:

I did not think the Wiki article said that.....can you explain where I am going wrong here:
Wikipedia:

From that I concluded the experiment showed NO directional deviation....negating Mach's idea...I seem to be missing something...[again!!]

Last edited by a moderator: May 6, 2017
6. Mar 8, 2013

### Vincentius

My question is not if Mach's principle is right, or Brans-Dicke theory, but instead why GR would not show anisotropic behavior. Radial length contraction is definitely anisotropic, so direction of the device must affect harmonic oscillator motion, I suspect. I am looking for an article on this subject. Has anyone a reference?

7. Mar 8, 2013

### Staff: Mentor

What do you mean by "radial length contraction"?

8. Mar 8, 2013

### Staff: Mentor

Vincentius, if you find a reference about the radial length contraction anisotropy then send me a PM and I will reopen the thread.

Last edited: Mar 8, 2013
9. Mar 8, 2013

### Staff: Mentor

Hi Everyone, as requested Vincentius sent a message to me with a reference describing what he meant by gravitational length contraction. Here are the references:
Please remember to keep the discussion within the forum rules, i.e. use mainstream scientific sources and avoid personal speculation.

Vincentius, the equation in the Google Books reference is isotropic. The equation in the second one I will need to go over in more detail, but it looks simply like the use of non-isotropic coordinates rather than any physical anisotropy.

Last edited: Mar 8, 2013
10. Mar 8, 2013

### Staff: Mentor

This is the way I read it as well. He appears to be using what we now call Schwarzschild coordinates, in which the radial metric coefficient $g_{rr}$ is different from the tangential coefficients $g_{\theta \theta}$ and $g_{\phi \phi}$. So the coordinates are non-isotropic; but that doesn't mean the unit length "measuring rods" he talks about are non-isotropic. Indeed, if you read carefully, you can see that he *assumes* that the measuring rods themselves *are the same length whether they are oriented radially or tangentially; that's how he derives the formulas in terms of the metric coefficients.

However, there *is* something "physical" going on here; but it's not anything to do with the measuring rods. It's to do with the geometry of space itself; the geometry of space as seen by an observer who is at rest in a static gravitational field is non-Euclidean. Einstein makes this clear in the paragraph following equation (71a).

11. Mar 8, 2013

### PAllen

I recall a thread on this maybe a year ago. There are a sprinkling of sources that discuss this. In that thread, in an attempt to clarify a measurable versus coordinate effect, I proposed the idea of rulers laid out from a planet viewed orthogonally to the rulers at distance. Would the rulers closer to the planet subtend less angle than spatially flat model would predict? Unfortunately, neither I nor other participants at the time were willing to do the somewhat laborious computations needed to answer that question.

Here are more links that supporting the possibility of this idea:

http://www.mth.uct.ac.za/omei/gr/chap8/node8.html

http://mathpages.com/rr/s6-01/6-01.htm (search for length contraction)

12. Mar 8, 2013

### bcrowell

Staff Emeritus
Look at the rest of the article where they explain that that original interpetation was wrong.

13. Mar 9, 2013

### PAllen

An additional comment on detecting length contraction: No observer detects anything but isotropy in their own local frame; this is obviously true even of the SR length contraction. It is a different observer who detects anisotropy in the affected object. Similarly for the hypothetical radial length contraction due to gravity - if it is observable it would be by distant observer as outlined in my scenario.

14. Mar 9, 2013

### Naty1

Ben,or anybody, for the life of me I cannot see where the Hughes-Drever experiments are said to fail to test Mach's principle....

This seems ok to me:
Vincentius posted:
to clarify, I am not questioning isotropy in SR/GR....just how the Hughes-Drever experiments failed to test Mach's principle....

These Hughes-Drever experiments showed the isotropy of nuclear resonance to a very accurate tolerance, right?? In other words, they found no physical relationship between the the distant stars and a local local inertial frame as Mach would have predicted.

I had never heard of these experiments before and so am interested if they really DO negate Mach's principle....apparently not, but I am stumped why not.

You posted
It's after the "but......" that I don't understand.

As I read the Wikipedia article, under MODERN INTERPRETATION, it says to me that after the Hughes-Drever experiments, it was realized the experiments ALSO applied to SR/GR:

thanks....

15. Mar 9, 2013

### Vincentius

I agree that locally everything is isotropic. And I also agree that proper spacetime near a mass is anisotropic in the eyes of a remote observer. Furthermore, relativistic trajectories like the perihelion precession are explained entirely by the anisotropy of the metric near a mass, so it is physical, not a coordinate artefact.
Then, if we have two identical harmonic oscillators on the same spot, one cycling in the radial direction of the massive body, the other one cycling in a perpendicular direction, then the two will run synchronously because of local isotropy (=Hughes-Drever). The remote observer, however, should be able to detect anisotropy in the motion of the oscillators, physically. But obviously, he still must see synchronous oscillators.
So, the question is: while anisotropy somehow affects the oscillators physically, why is it not visible in the cycle time? Hence, anisotropy either does not affect cycle time, or the effect is canceled by something else. This is what I like to get clearified. I would expect this has been explained times and over again, but I can not find a reference. Anyone knows? I appreciate your contribution.

16. Mar 9, 2013

### Staff: Mentor

Excellent.

Huh? How can you "agree that X" when nobody has said X. Furthermore, "proper spacetime" is not a standard term. I don't know what you are talking about. If you wish to discuss "proper spacetime" please provide a reference.

No. You can use isotropic coordinates and get the exact same precessing geodesics:

http://en.wikipedia.org/wiki/Schwar...c.29_formulations_of_the_Schwarzschild_metric

So anisotropy in the Schwarzschild metric is a coordinate artifact and is not necessary to explain the geodesics.

How?

We certainly haven't established that it does affect oscillators anisotropically in any coordinate-indepentent fashion.

17. Mar 9, 2013

### bcrowell

Staff Emeritus
Regardless of whether WP says so explicitly, the original interpretation was wrong, for the reasons I gave in #4.

18. Mar 10, 2013

### Vincentius

I agreed with PAllen #13. And I agree with you that "proper spacetime" is confusing. Just delete "proper".

I can not imagine that one can derive perihelion precession from isotropic coordinates (that is, without converting back to anisotropic coordinates of course). Can you give a reference DaleSpam?

Last edited by a moderator: May 6, 2017
19. Mar 10, 2013

### PAllen

It is essentially self evident that you can. Perihelion advance is a feature of shift in the angular coordinate of closest approach. Between isotropic SC coordinates and regular SC coordinates, the angular coordinates don't change (neither does the time coordinate). Only the radial is scaled. Thus, the transform of an orbit exhibiting with perihelion advance to isotropic coordinates shows exactly same advance, in exactly the same form. It remains a geodesic of the metric expressed in isotropic coordinates.

20. Mar 10, 2013

### Staff: Mentor

Sure. See Carrol's lecture notes, p 55-58 derives the covariant derivative to "perform the functions of the partial derivative, but in a way independent of coordinates". Then equation 3.30 and 3.46 define a geodesic in terms of the covariant derivative.

http://arxiv.org/abs/gr-qc/9712019

Since a geodesic is defined in terms of the covariant derivative, and since the covariant derivative is independent of coordinates, then if a path is a geodesic in anisotropic coordinates it remains a geodesic in any other coordinates, including isotropic coordinates. I.e. the orbit is the same orbit regardless of the coordinates.

Last edited: Mar 10, 2013